The K.E of N molecules of `O_(2)` is x Joules at -123°C. Another sample of `O_(2)` at 27°C has a KE of 2x Joules. The latter sample contains________N molecules of `O_(2)`.

Answer

To solve the problem, we need to use the kinetic energy formula for gases and analyze the given information step by step.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The first sample of \( O_2 \) is at -123°C.
- To convert to Kelvin:
\[ T_1 = -123 + 273 = 150 \, K \]
- The second sample is at 27°C.
- To convert to Kelvin:
\[ T_2 = 27 + 273 = 300 \, K \]

2. **Write the Kinetic Energy Equation**:
- The kinetic energy (KE) of a gas is given by the formula:
\[ KE = \frac{3}{2} nRT \]
- For the first sample:
\[ KE_1 = \frac{3}{2} n R T_1 = \frac{3}{2} n R (150) \]
- For the second sample:
\[ KE_2 = \frac{3}{2} y R T_2 = \frac{3}{2} y R (300) \]
- Here, \( n \) is the number of molecules in the first sample, and \( y \) is the number of molecules in the second sample.

3. **Set Up the Equations**:
- We know from the problem that:
\[ KE_1 = x \quad \text{and} \quad KE_2 = 2x \]
- Therefore, we can write:
\[ \frac{3}{2} n R (150) = x \quad \text{(1)} \]
\[ \frac{3}{2} y R (300) = 2x \quad \text{(2)} \]

4. **Relate the Two Equations**:
- From equation (1):
\[ x = \frac{3}{2} n R (150) \]
- From equation (2):
\[ 2x = \frac{3}{2} y R (300) \]
- Substitute \( x \) from equation (1) into equation (2):
\[ 2 \left( \frac{3}{2} n R (150) \right) = \frac{3}{2} y R (300) \]

5. **Simplify the Equation**:
- Cancel \( \frac{3}{2} R \) from both sides:
\[ 2n(150) = y(300) \]
- Simplifying gives:
\[ 300n = 300y \]
- Therefore, we find:
\[ y = n \]

6. **Conclusion**:
- The second sample contains \( n \) molecules of \( O_2 \).

### Final Answer:
The latter sample contains **N molecules of \( O_2 \)**.